Flow through a hole in a sphere

I have a sphere of diameter $$D$$ (radius $$R$$) with a small hole of diameter $$d$$ (radius $$a$$) in it, with air flowing through the axis of the hole:

I am trying to estimate the drag force due to friction in the hole.

First let's consider simplifying assumptions:

1. $$d << D$$
2. Pressure distribution around the sphere is unaffected by the small hole.
3. $$Re_D = \frac{\rho u_{\infty}D}{\mu} >> 1 \therefore$$ inviscid flow around sphere.
4. $$Re_d\frac{d}{D} << 1 \therefore$$ inertia free flow in the hole via Navier-Stokes scaling.

Assumptions 2 and 3 suggest that pressures at points 1 and 2 (shown in the figure) can be estimated via simple stagnation pressures with Bernoulli's equation:

$$P_1 = P_2 = P_{\infty} + \frac{1}{2} \rho u_{\infty}^2$$

Now, assumptions 1 and 4 suggest that the flow in the hole is viscous dominated, thus yielding the Hagen-Poiseuille solution for flow through a pipe:

$$v_z(r) = \frac{1}{4\mu}\frac{\Delta P}{D}(a^2 - r^2)$$

From this, I could easily calculate the shear stress in the hole, and therefore the drag force in the hole.

The problem, however, is that my Bernoulli analysis yields a zero pressure drop:

$$\Delta P = P_1 - P_2 = 0$$

This seems to suggest that there is no flow through the small hole in an inviscid regime, where $$Re_D >> 1$$.

In that case, I have zero friction drag in the hole since there is approximately no flow.

Is this analysis correct? Would there be approximately no flow and therefore no friction drag in this small hole, if there is large Reynolds number flow around the sphere?

• Even at high Re, I still think there would be a significant pressure difference between 1 and 2. This is because of the no slip boundary condition on the sphere, and the separation zone that develops on trailing edge of the sphere. Look up the flow distribution past a solid sphere as Re increases. – Chet Miller Nov 29 '18 at 0:13
• ecourses.ou.edu/cgi-bin/… – Chet Miller Nov 29 '18 at 0:28
• As a crude first approximation I think you can assume that the fluid behind the sphere (at rear stagnation point) is stationary (because of recirculating region there) so that the pressure difference $p_1-p_2\sim 0.5\rho u_\infty^2$. – Deep Nov 29 '18 at 6:11
• @ChesterMiller you're right, there is a definite pressure drop. My purely inviscid analysis is incorrect and my mistake actually looks a lot like D'Alembert's paradox. – Drew Nov 29 '18 at 14:08
• @Deep This is what I've seen from other sources. I've tried to analytically show that $\Delta P \sim 0.5\rho u_{\infty}^2$ in my answer below, although I'm not sure if it's a proper argument. – Drew Nov 29 '18 at 14:40

It looks like my issue of $$\Delta P = 0$$ is a case of D'Alembert's paradox, where purely inviscid flow wrongfully predicts zero pressure drop around an object. Furthermore, if this is high Reynolds number flow $$(Re_D >> 1 )$$, it is most likely in the turbulent regime, and experiments have shown that that $$\Delta P \sim 0.5\rho u_{\infty}^2$$, as shown below with the coefficient of pressure $$C_P$$.

This result was suggested by user Deep in a comment on the OP. Here I will attempt an argument to show this result analytically.

My analysis for the stagnation pressure at point 1 is fine:

$$P_1 = P_{\infty} + 0.5\rho u_{\infty}^2$$

Now we can consider the drag force and some simple scaling to arrive at $$P_2$$.

For a sphere in the high $$Re_D$$ regime, we know that the drag coefficient $$C_D = 0.4$$, thus the drag force is:

$$F_D = 0.4(0.5\rho u_{\infty}^2)\pi R^2 = 0.628\rho u_{\infty}^2R^2$$

And $$F_D \sim R^2\Delta P$$, therefore:

$$\Delta P = P_1 - P_2 \sim 0.628\rho u_{\infty}^2$$

Solving for $$P_2$$,

$$P_2 \sim P_1 - 0.628\rho u_{\infty}^2 = P_{\infty} - 0.128\rho u_{\infty}^2$$

where we can ignore the $$0.128\rho u_{\infty}^2$$ term? If we do ignore it, the result is:

$$P_2 \sim P_{\infty}$$

This seems kind of hand-wavy, but it's the closest I've come to convincing myself with analytical arguments that $$P_2 \sim P_{\infty}$$. Does anyone know of a better analysis?

• When Reynolds number becomes high enough we only get hand-waving arguments. In high Re flows, drag is primarily due to pressure drop rather than skin friction. So your order-of-magnitude analysis is good enough. – Deep Nov 29 '18 at 14:43